CBSE X Maths Constructions

Construction of a tangent to a circle (using the centre)
We remember that: In a circle, the radius drawn at the point of contact is perpendicular to the tangent at
that point.
Q. Draw a circle of radius 3.2 cm. Take a point P on this circle and draw a tangent at P.(using the centre)   Given: Radius of the circle = 3.2 cm.

Steps of construction

(i) With O as the centre draw a circle of radius 3.2 cm.
(ii) Take a point P on the circle and join OP.
(iii) Draw an arc of a circle with centre at P cutting OP at L.
(iv) Mark M and N on the arc such that m(arc L M ) = m(arcMN)
(v) Draw the bisector PT of the angle <MPN
(vi) Produce TP to Tlto get the required tangent T'PT.
Construction of pair of tangents to a circle from an external point

We remember that:

(i) Two tangents can be drawn to a circle from an external point.
(ii) Diameters subtend 90 degree on the circumference of a circle.

Q. Draw a circle of radius 3 cm. From an external point 7 cm away from its centre, construct
the pair of tangents to the circle and measure their lengths.
Given: Radius of the circle = 3 cm. OP = 7 cm.
Construction Steps:
(i) With O as the centre draw a circle of radius 3 cm.
(ii) Mark a point P at a distance of 7 cm from O and join OP.
(iii) Draw the perpendicular bisector of OP. Let it meet OP at M.
(iv) With M as centre and MO as radius, draw another circle.
(v) Let the two circles intersect at T and Tl.
(vi) Join PT and PT'. They are the required tangents.
Length of the tangent, PT = 6.3 cm
Q. Draw a line segment of length 7 cm and divide it in the ratio 2 : 3.
Solution. Step of constructions :
1. Draw AB = 7 cm
2. Draw ray AX making a suitable acute angle with AB.
3. Cut 2 + 3 = 5 equal segments AA1, A1A2, A2A3, A3A4 and A4A5 on AX.
4. Join A5 with B.

5. Through A2 ,Draw A2P parallel to A5B by making corresponding angles AA2P and AA5B
equal.
6. The line through A2 and parallel to A5B will meet the given line segment at point P.
Then P is the required point which divides AB in the ratio 2 : 3 i.e. AP : PB = 2 : 3
Q. Construct a triangle similar to a given triangle with sides 7 cm, 9 cm and 10 cm and whose sides are 5/7 th of the corresponding sides of the given triangle.
Solution.
Steps of Constructions :
1. With the given measurements construct the triangle ABC in which AB = 7 cm, BC = 9 cm andAC = 10 cm
2. Draw a ray AP, making any suitable angle with AB and on opposite side of vertex C
3. Starting from A, cut off seven equal line-segments AX1, X1X2, X2X3, X3X4, X4X5, X5X6 and X6X7 on AP.
4. Join BX1 and draw a line X5B' parallel to X7B which meets AB at B'.
5. Through B', draw B'C' || BC which meets AC at point C'.

The DAB'C', so obtained, is similar to the given D ABC and each side of DAB'C' is 5/7 times the corresponding side of D ABC.

Q. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given
triangle.
Solution.
 Step of Constructions :
1. Draw BC = 4 cm
2. At B, draw a ray BP making angle 90° with BC i.e. <PBC = 90°
3. From BP, cut BA = 3 cm
4. Join A and C to get the given DABC
5. Through vertex B, draw ray BX making any suitable angle with BC.
6. On BX cut 5 equal line segments BB1 = B1B2 = B2B3 = B3B4 = B4B5.
7. Join B3 to C. 

8. Through B5, draw a line parallel to B3C to meet BC produced at point C'.
9. Through C', draw a line parallel to side CA to meet BA produced to A'.

D A'BC' is the required triangle